An object with a mass of 8 kg is hanging from a spring with a constant of 9 kgs^-2. If the spring is stretched by  12 m, what is the net force on the object?

Jan 18, 2016

Total upward restoring force of the spring: $108 N$.
Total downward gravitational force on the mass: $78.4 N$
Net force on the mass: $29.6 N u p w a r d$

Explanation:

Kilogram per square second $\left(k g {s}^{-} 2\right)$ is an odd unit for the spring constant, $k$, of a spring. This is usually expressed in terms of newton per metre $\left(N {m}^{-} 1\right)$ - for every additional metre a spring is stretched, it exerts a force of $k$ newton.

The two units are actually the same, as a dimensional analysis will show. The newton, $N$, is defined as a $k g m {s}^{-} 2$. If we multiply this by ${m}^{-} 1$ the $m$ just cancels out and and we are left with $k g {s}^{-} 2$.

Why am I making a big deal of this? Because this question makes a lot more sense with the correct units:

"An object with a mass of $8 k g$ is hanging from a spring with a constant of $9 N {m}^{-} 1$. If the spring is stretched by $12 m$, what is the net force on the object?"

For each metre the spring is stretched, the force is increased by $9 N$, so a $12 m$ stretch leads to a $12 \cdot 9 = 108 N$ increase in the force.

This is an upward restoring force (back toward the unstretched center position of the spring), but to calculate the net force we also need to take into account the downward force of gravity on the mass:

$F = m g = 8 \cdot 9.8 = 78.4 N$

Since these forces are acting in opposite directions we subtract one from the other, and end up with a net upward force of $29.6 N$.